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3 years ago

The chart shows the bids provided by four engineers to test a prototype.

Engineering

1 answer:

klasskru [66]3 years ago

8 0

Answer:

Explanation:

To know which is most or least cost-effective, it's not enough to look at only the per day rate, or only the time to complete. You have to multiply them to get the total cost of the project.

$\left[\begin{array}{ccccc}&Cost\ per\ day\ (\$)&Time\ to\ complete\ (days)&Total\ cost\ (\$)\\Zoe&500&8&4000\\Greg&650&10&6500\\Orion&400&12&4800\\Jin&700&5&3500\end{array}\right]$

As you can see, Greg is the least cost-effective because he charges the most for the project.

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The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr

djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

6 0

3 years ago

How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?

Digiron [165]

Answer:

Two stroke cycle Four stroke cycle

1.Have on power stroke in one revolution. 1.have one power

stroke in two revolution

2.Complete the cycle in 2 stroke 2.Complete the cycle in 4 stroke

3.It have ports 3.It have vales

4.Greater requirement of cooling 4.Lesser requirement of cooling

5.Less thermal efficiency 5.High thermal efficiency

6.Less volumetric efficiency 6.High volumetric efficiency

7.Size of flywheel is less. 7.Size of flywheel is more.

3 0

3 years ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

mr_godi [17]

Answer:

-50.005 KJ

Explanation:

Mass flow rate = 0.147 KJ per kg

mass= 10 kg

Δh= 50 m

Δv= 15 m/s

W= 10×0.147= 1.47 KJ

Δu= -5 kJ/kg

ΔKE + ΔPE+ ΔU= Q-W

0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W

Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu

= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50

= 1.47 +3.375-4.8450-50

Q=-50.005 KJ

7 0

3 years ago

Read 2 more answers

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

6 0

3 years ago

state & prove parallelogram law of vector addition &Also determine magnitude &direction of resultant vector.

ludmilkaskok [199]

Answer:

Parallelogram law of vector addition states that if two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of the two vectors is given by the vector that is diagonal passing through the point of contact of two vectors.

8 0

3 years ago